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shtirl [24]
3 years ago
13

1) The mean of eight numbers is 75. If the mean of three of the numbers is 60, find the mean of the remaining five numbers

Mathematics
1 answer:
Ymorist [56]3 years ago
4 0
1.
mean=average of all them


(a+b+c+d+e+f+g+h)/8=75
a+b+c+d+e+f+g+h=600


(a+b+c)/3=60
a+b+c=180
subsitute

a+b+c+d+e+f+g+h=600
180+d+e+f+g+h=600
minus 180 both sides
d+e+f+g+h=420
divvide both side by 5
(d+e+f+g+h)/5=84
the mean is 84




2.
1/3 are increased by 6
that means 30 of them are increased by 6
30*6=180
adds 180 to total
180/90=2
increases mean by 2
47+2=49
new mean is 49



c.
v=hpir^2
h=50ft, or 12*50in or 600in
r=1
v=600pi1^2
v=600pi
use 3.14
v=1884 cubic inches

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Which of the following theorems verifies that ABC~WXY?
Bumek [7]

Answer:

The correct answer is option <em>B. AA</em>

<em></em>

Step-by-step explanation:

Given two triangle:

\triangle ABC  and \triangle WXY.

The dimensions given in \triangle ABC are:

\angle A = 27^\circ\\\angle B = 90^\circ

We know that the sum of three angles in a triangle is equal to 180^\circ.

\angle A+\angle B+\angle C = 180^\circ\\\Rightarrow 27+90+\angle C=180^\circ\\\Rightarrow \angle C = 63^\circ

The dimensions given in \triangle WXY are:

\angle Y = 63^\circ\\\angle X = 90^\circ

We know that the sum of three angles in a triangle is equal to 180^\circ.

\angle W+\angle X+\angle Y = 180^\circ\\\Rightarrow \angle W+90+63=180^\circ\\\Rightarrow \angle W = 27^\circ

Now, if we compare the angles of the two triangles:

\angle A = \angle W = 27^\circ\\\angle B = \angle X= 90^\circ\\\angle C = \angle Y= 63^\circ

So, by AA postulate (i.e. Angle - Angle) postulate, the two triangles are similar.

\triangle ABC \sim \triangle WXY by <em>AA theorem.</em>

<em>So, correct answer is option B. AA </em>

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3 years ago
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